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Setup - Linda is bored and decides to pour an entire container of salt into a pile on the kitchen floor. She pours $3$ cubic inches of salt per second into a conical pile whose height is always two-thirds of its radius.

Question - How fast is the radius of the conical salt pile changing when the radius of the pile is $2$ inches?

So I don't really know what to do here. Could someone walk me through it? I'm assuming you use the volume and surface area formula of a cone.

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First, let $t$ be time in seconds, $r(t)$ be the radius in inches at time $t$ seconds, $h(t)$ be the height in inches at time $t$ seconds, and $V(t)$ be the volume in cubic inches at time $t$ seconds, and let $t_0$ be the time at which the radius is $2$ inches.

What do we know? The volume is increasing at a rate of $3$ cubic inches per second, so $$\frac{\text dV}{\text dt}(t) = 3$$ for all $t$. Height is two-thirds the radius, so $$h(t) = \frac{2}{3}r(t)$$ for all $t$. The radius is $2$ inches at time $t_0$, so $$r(t_0) = 2$$

What do we want? We are asked how fast the radius is changing (with respect to time) at the instant where the radius is $2$ inches, so we want to find $\frac{\text dr}{\text dt}(t_0)$.

What do we need? We have a relationship between $h$ and $r$, but in terms of rates, we only have information about $\frac{\text dV}{\text dt}$. So, we need some relationship between $V$ and $r$, $h$, or both. Since the shape is a cone, we can can use the formula for the volume of a cone, $$V(t) = \frac{1}{3}\pi h(t)r(t)^2$$ Substituting our relationship between $r$ and $h$, we get $$V(t) = \frac{2}{9}\pi r(t)^3$$ Differentiating this gives us $$\frac{\text dV}{\text dt}(t) = \frac{2}{3}\pi r(t)^2\frac{\text dr}{\text dt}(t)$$

Now that we have all the information, we just need to put it together. Plugging in $t=t_0$ will give us the relationship when the radius is $2$ inches, $$\frac{\text dV}{\text dt}(t_0) = \frac{2}{3}\pi r(t_0)^2\frac{\text dr}{\text dt}(t_0)$$ However, we know that $\frac{\text dV}{\text dt}(t) = 3$ for all $t$, and in particular at $t=t_0$, so we have $$\frac{\text dV}{\text dt}(t_0) = 3$$ We also know that $r(t_0) = 2$ by definition of $t_0$. This means that our equation reduces to $$3 = \frac{2}{3}\pi\cdot 2^2 \frac{\text dr}{\text dt}(t_0)$$ which we can solve to determine the value $$\frac{\text dr}{\text dt}(t_0) = \frac{9}{8}\pi$$

In terms of the original problem, $\frac{\text dr}{\text dt}(t_0) = \frac{9}{8}\pi$ can be interpreted as the salt pile's radius increasing at a rate of $\frac{9}{8}\pi$ inches per second when the radius is $2$ inches.

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We only have to use the formula for volume.

We know two things: $V=\frac13\pi r^2h$ and $h=\frac23r$.

Let's sub this in to get the volume, $V$, as a function of the radius, $r$.

$V(r) = \frac29\pi r^3$

Next, we derive this function with respect to time, $t$.

$\frac{dV}{dt} = \frac23\pi r^2 \frac{dr}{dt}$

We were given a few values in the setup: $\frac{dV}{dt}=3$, that is, the volume of the pile is increasing at 3 in$^3/$sec. The radius of the pile is 2 in, and we need to find $\frac{dr}{dt}$, how fast the radius is changing when $r=2$.

$$3=\frac23\pi\cdot4\cdot\frac{dr}{dt}$$ $$\frac{dr}{dt} = \frac{9\pi}8$$

The radius of the pile is changing at a rate of $\frac{9\pi}8$ in/sec.

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